如何发送包含嵌入图片的HTML电子邮件? HTML应该如何链接到图像?图像应作为MultiPart电子邮件附件添加?
非常感谢任何一个例子。
答案 0 :(得分:8)
http://djangosnippets.org/snippets/285/
您必须使用MultiPart和cid:发送带图像的HTML邮件几乎总是一个坏主意。它为您的邮件和smtp服务器提供垃圾邮件点......
答案 1 :(得分:6)
请记住,django只提供标准smtplib的包装 - 我不知道它是否会有所帮助,但请尝试查看此示例:http://code.activestate.com/recipes/473810-send-an-html-email-with-embedded-image-and-plain-t/
所以我猜你可以使用EmailMessage
的标题值来定义这个'image1' - 消息标题是值的字典,所以只需添加类似{'Content-ID': '<image1>'}
的内容。
然后使用attach()
将文件附加到您的电子邮件中。之后,您可以使用代码生成如下的html消息:
html_content = '<b>Some HTML text</b> and an image: <img src="cid:image1">'
答案 2 :(得分:4)
我实现了op要求使用django的邮件系统。它会增加它将使用django设置进行邮件发送(包括用于测试的不同子系统等等。我还在开发期间使用mailhog)。这也是一个更高的水平:
from django.conf import settings
from django.core.mail import EmailMultiAlternatives
message = EmailMultiAlternatives(
subject=subject,
body=body_text,
from_email=settings.DEFAULT_FROM_EMAIL,
to=recipients,
**kwargs
)
message.mixed_subtype = 'related'
message.attach_alternative(body_html, "text/html")
message.attach(logo_data())
message.send(fail_silently=False)
logo_data
是一个附加徽标的辅助函数(在这种情况下我想附加的图像):
from email.mime.image import MIMEImage
from django.contrib.staticfiles import finders
@lru_cache()
def logo_data():
with open(finders.find('emails/logo.png'), 'rb') as f:
logo_data = f.read()
logo = MIMEImage(logo_data)
logo.add_header('Content-ID', '<logo>')
return logo
答案 3 :(得分:0)
如果您要发送带有图像作为附件的电子邮件(在我的情况下,这是保存在表单中的图像,直接保存下来),则可以使用以下代码作为示例:
#forms.py
from django import forms
from django.core.mail import EmailMessage
from email.mime.image import MIMEImage
class MyForm(forms.Form):
#...
def save(self, *args, **kwargs):
# In next line we save all data from form as usual.
super(MyForm, self).save(*args, **kwargs)
#...
# Your additional post_save login can be here.
#...
# In my case name of field was an "image".
image = self.cleaned_data.get('image', None)
# Then we create an "EmailMessage" object as usual.
msg = EmailMessage(
'Hello',
'Body goes here',
'from@example.com',
['to1@example.com', 'to2@example.com'],
['bcc@example.com'],
reply_to=['another@example.com'],
headers={'Message-ID': 'foo'},
)
# Then set "html" as default content subtype.
msg.content_subtype = "html"
# If there is an image, let's attach it to message.
if image:
mime_image = MIMEImage(image.read())
mime_image.add_header('Content-ID', '<image>')
msg.attach(mime_image)
# Then we send message.
msg.send()
答案 4 :(得分:0)
我已经尝试了以下代码,并且可以正常工作。
代码:
msg = EmailMessage()
# generic email headers
msg['Subject'] = 'Welcome'
msg['From'] = 'abc@gmail.com'
recipients = ['abc@gmail.com']
# set the plain text body
msg.set_content('This is a plain text body.')
# now create a Content-ID for the image
image_cid = make_msgid(domain='')
# if `domain` argument isn't provided, it will
# use your computer's name
# set an alternative html body
msg.add_alternative("""\
<html>
<body>
<table border='0' cellpadding='1' cellspacing='0' width='800'>
<tbody>
<tr>
<td height='506'>
<table border='0' cellpadding='0' cellspacing='0' width='600'>
<tbody>
<tr>
<td valign='top'>
<img height='190' src="cid:{image_cid}" width='800' tabindex='0'>
</td>
</tr>
<tr>
<td height='306' valign='top'>
<table cellpadding='0' cellspacing='20' width='800'>
<tbody>
<tr>
<td align='left' height='804' style='font-family:arial,helvetica,sans-serif;font-size:13px' valign='top'>
Hi {name},<br><br>
Welcome!
</td>
</tr>
</tbody>
</table>
</td>
</tr>
</tbody>
</table>
</td>
</tr>
</tbody>
</table>
</body>
</html>
""".format(image_cid=image_cid[1:-1],name='ABC'), subtype='html')
# image_cid looks like <long.random.number@xyz.com>
# to use it as the img src, we don't need `<` or `>`
# so we use [1:-1] to strip them off
# now open the image and attach it to the email
with open('/path/image.jpg', 'rb') as img:
# know the Content-Type of the image
maintype, subtype = mimetypes.guess_type(img.name)[0].split('/')
# attach it
msg.get_payload()[1].add_related(img.read(),
maintype=maintype,
subtype=subtype,
cid=image_cid)
server = smtplib.SMTP(host=<hostname>, port=25)
server.starttls()
# send the message via the server.
server.sendmail(msg['From'], recipients, msg.as_string())
server.quit()